Science:Math Exam Resources/Courses/MATH110/December 2017/Question 03 (b)

MATH110 December 2017

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Question 03 (b)
Evaluate the following limits or determine they are either infinite or do not exist. (b) ${\textstyle \displaystyle \lim _{x\to -2^{+}}{\frac {x^{2}+3x+2}{x^{2}+5x+6}}}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
What are the limits of the numerator and the denominator, respectively?

Hint 2
Factorize and look for cancellations.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let ${\textstyle f(x)=x^{2}+3x+2}$ and ${\textstyle g(x)=x^{2}+5x+6}$. Since they are (quadratic) polynomials and are therefore continuous everywhere, we compute $$\lim _{x\to -2^{+}}f(x)=(-2)^{2}+3(-2)+2=4-6+2=0$$ and $$\lim _{x\to -2^{+}}g(x)=(-2)^{2}+5(-2)+6=4-10+6=0.$$ This means that both ${\textstyle f(x)}$ and ${\textstyle g(x)}$ carry a common factor of ${\textstyle x+2}$. In order to cancel such factor, we write $$f(x)=(x+1)(x+2)$$ and $$g(x)=(x+2)(x+3).$$ Then for any real number ${\textstyle x\neq -2}$, we have $${\dfrac {f(x)}{g(x)}}={\dfrac {(x+1)(x+2)}{(x+2)(x+3)}}={\dfrac {x+1}{x+3}}.$$ In particular, the given limit is $$\lim _{x\to -2^{+}}{\dfrac {f(x)}{g(x)}}=\lim _{x\to -2^{+}}{\dfrac {x+1}{x+3}}={\dfrac {-2+1}{-2+3}}={\dfrac {-1}{1}}=-1.$$ Answer: ${\textstyle \lim \limits _{x\to -2^{+}}{\dfrac {x^{2}+3x+2}{x^{2}+5x+6}}={\color {blue}-1}}$.